Problem: In the diagram, $ABCD$ is a square with side length $6,$ and $WXYZ$ is a rectangle with $ZY=10$ and $XY=6.$ Also, $AD$ and $WX$ are perpendicular. If the shaded area is equal to half of the area of $WXYZ,$ what is the length of $AP?$

[asy]
draw((0,0)--(10,0)--(10,6)--(0,6)--cycle,black+linewidth(1));
draw((1.5,1)--(7.5,1)--(7.5,7)--(1.5,7)--cycle,black+linewidth(1));
filldraw((1.5,1)--(7.5,1)--(7.5,6)--(1.5,6)--cycle,gray,black+linewidth(1));
label("$W$",(0,6),NW);
label("$X$",(10,6),NE);
label("$Y$",(10,0),SE);
label("$Z$",(0,0),SW);
label("$A$",(1.5,7),NW);
label("$B$",(7.5,7),NE);
label("$C$",(7.5,1),E);
label("$D$",(1.5,1),W);
label("$P$",(1.5,6),SW);
label("6",(1.5,7)--(7.5,7),N);
label("6",(10,0)--(10,6),E);
label("10",(0,0)--(10,0),S);
[/asy]
The area of rectangle $WXYZ$ is $10 \times 6 = 60.$

Since the shaded area is half of the total area of $WXYZ,$ its area is $\frac{1}{2}\times 60=30.$

Since $AD$ and $WX$ are perpendicular, the shaded area has four right angles, so is a rectangle.

Since square $ABCD$ has a side length of $6,$ we have $DC=6.$

Since the shaded area is $30,$ then $PD \times DC=30$ or $PD \times 6 = 30$ or $PD=5.$

Since $AD=6$ and $PD=5,$ we obtain $AP=\boxed{1}.$